Chemistry
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§ 06.06.a1
Energy: Standard Enthalpy of Formation
Date:
& Reaction
45 min
Hess’s Law
Most Important Ideas:
 When reactants are converted to products, the change in enthalpy is the same whether the
reaction takes place in one step or in a series of steps.
References
1. Textbook pages 199-201, including Exercises & Practice 6.5 & 6.6.
2. YouTube: IsaacTeach: Chemistry 10.7b Hess's Law of Heat Summation
(http://www.youtube.com/watch?v=Su44G7-NvKE).
Background
The enthalpy of formation of a substance
indicates how stable it is. One way to determine
is by the direct method is to use the chemical reaction and apply
= (sum of
for the
products) – (sum of
for the reactants). Another important way, and more commonly used,
is Hess’s law. Because
is a state function, only the initial and final values are important; it
is not necessary to know what was done in between. This allows us the take a series of steps,
using published values, to determine
or
.
Model
Problem
Calculate the standard enthalpy for the reaction of N2(g) + 2O2(g)  2NO2(g).
Solution.
We look up (or are given, as in class worksheets or on a quiz or test)
for each of the
substances:
(a)
N2(g) +
1O2(g)

2NO2(g) 
(b)
2NO(g)
= +180 kJ
2NO(g) + O2(g)
= +112 kJ
1. Label the intermediary reactions (a), (b), etc. This helps a lot in not getting confused with all
of the steps you may need to use. (The ‘a’ and ‘b’ are done above but may not be given
explicitly to you in the problem.)
2. Align the order of each step to match substances as reactant or product. The second step
needed to be reversed so that the NO2 is now a product. Notice that the sign of
for this
step is now a negative.
N2(g)
+
2O2(g)

2NO2(g)
=
(a)
N2(g)
+
O2(g)
2NO(g)
= +180 kJ
(–b)
2NO(g) +
O2(g)


2NO2(g)
= –112 kJ
F:\330\330_sections\330_06_Energy\330_06_06_Standard Enthalpy_Formation_Reaction\330_06_06b1_Standard_Enthalpy_Hess_Law_HW.docx (3/21/2014 1:18:00 PM)
? kJ
Chemistry
§06.06.b1 Energy: Heat of Formation – Hess’s Law
p. 2
3. It is often necessary to double or triple a reaction so that where the same substance appears
on both sides of the equation, it has the same number of moles. In this case we don’t need
to.
4. Add the steps together and cancel out substances that appear on opposite sides. (It is an
algebraic problem and the rules of math apply.)
N2(g)
+
2O2(g)

2NO2(g)
=
(a)
N2(g)
+
O2(g)
2NO(g)
= +180 kJ
(–b)
2NO(g) +
O2(g)


2NO2(g)
= –112 kJ

2NO2(g)
= +68 kJ
5. Rewrite the equation and solve for
N2(g)
+
2O2(g)
? kJ
.
Problems
6.61 From these data,
S(rhombic)
+ O2(g) → SO2(g)
= –296.06 kJ/mol
S(monoclinic) + O2(g) → SO2(g)
= –296.36 kJ/mol
Calculate the enthalpy change for the transformation:
S(rhombic) → S(monoclinic)
(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)
6.62 From the following data,
C(graphite) + O2(g) → CO2(g)
= –393.5 kJ/mol
H2(g) + ½O2(g) → H2O(l)
2C2H6(l) + 7O2(g) → 4CO2(g) + 6H2O(l)
Calculate the enthalpy change for the reaction:
2C(graphite) + 3H2(g) → 1C2H6 (g)
6.63 From the following heats of combustion,
CH3OH(l) + O2(g) → CO2(g) + 2H2O(l)
C(graphite) + O2(g) → CO2(g)
= –285.5 kJ/mol
= − 3119.6 kJ/mol
= − 726.4 kJ/mol
= –393.5 kJ/mol
H2(g) + ½O2(g) → H2O(l)
= –285.5 kJ/mol
Calculate the enthalpy of formation of methanol (CH3OH) from its elements:
C(graphite) + 2H2(g) + ½O2(g) → CH3OH(l)
6.64 Calculate the standard enthalpy change for the reaction:
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
given that
2Al(s) + O2(g) → Al2O3(s)
= –1669.8 kJ/mol
2Fe(s) + O2(g) → Fe2O3(s)
= –822.2 kJ/mol
Chemistry
§06.06.b1 Energy: Heat of Formation – Hess’s Law
p. 3
Extensions
6.73 Calculate the work (J) done by the reaction: 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
when 0.34 g of Na reacts with water to form hydrogen gas at 0oC and at 1.0 atm.
6.74 You are given the following data:
= +436.4 kJ/mol
H2(g)  2H(g)
(a)
(b)
(c)
Calculate
H2(g) +
Br2(g) 
Br2(g) 
for the reaction:
2Br(g)
2HBr(g)
H(g) + Br(g)  HBr(s)
= +192.5 kJ/mol
= –72.4 kJ/mol